Question

Let the two focii of an ellipse be (-1,0) and (3,4) and the foot of perpendicular from focus (3,4) upon a tangent to the ellipse be (4,6). Find (i) length of semi minor axis, (ii) foot of $\perp$ from focus (-1,0) upon the same tangent. (iii) equation of the auxiliary circle of ellipse. (iv) equations of the directrix.

• A. (i) $\sqrt{17}$, (ii) $(\frac{12}{5}, \frac{34}{5})$, (iii) $(x-1)^2 + (y-2)^2 = 25$, (iv) $x + y - 10 = 0$ and $5x + 5y - 46 = 0$
• B. (i) $\sqrt{17}$, (ii) $(\frac{12}{5}, \frac{34}{5})$, (iii) $(x-1)^2 + (y-2)^2 = 25$, (iv) $x + y - 10 = 0$ and $5x + 5y - 46 = 0$
• C. (i) $\sqrt{17}$, (ii) $(\frac{12}{5}, \frac{34}{5})$, (iii) $(x-1)^2 + (y-2)^2 = 25$, (iv) $x + y - 10 = 0$ and $5x + 5y - 46 = 0$
• D. (i) $\sqrt{17}$, (ii) $(\frac{12}{5}, \frac{34}{5})$, (iii) $(x-1)^2 + (y-2)^2 = 25$, (iv) $x + y - 10 = 0$ and $5x + 5y - 46 = 0$

Answer 1

Answer: (A), (B), (C), (D)

(i) Length of semi-minor axis: $\sqrt{17}$ (ii) Foot of $\perp$ from focus $(-1,0)$ upon the same tangent: $(\frac{12}{5}, \frac{34}{5})$ (iii) Equation of the auxiliary circle: $(x-1)^2 + (y-2)^2 = 25$ (iv) Equations of the directrices: $x + y - 10 = 0$ and $5x + 5y - 46 = 0$.

Answer 2

1. Center of Ellipse: The center $C$ is the midpoint of the foci $F_1(-1, 0)$ and $F_2(3, 4)$. $C = \left(\frac{-1+3}{2}, \frac{0+4}{2}\right) = (1, 2)$.

2. Distance between Foci (2c): $2c = \sqrt{(3 - (-1))^2 + (4 - 0)^2} = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}$. So, $c = 2\sqrt{2}$.

3. Semi-major Axis (a): The foot of the perpendicular from a focus to a tangent lies on the auxiliary circle. Thus, $P(4, 6)$ lies on the auxiliary circle centered at $C(1, 2)$. The radius of the auxiliary circle is $a$. $a = CP = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.

4. (i) Length of Semi-minor Axis (b): Using $a^2 = b^2 + c^2$: $5^2 = b^2 + (2\sqrt{2})^2$ $25 = b^2 + 8$ $b^2 = 17 \implies b = \sqrt{17}$.

5. (ii) Foot of $\perp$ from $F_1$ to the tangent: The slope of $F_2P$ is $m_{F_2P} = \frac{6-4}{4-3} = 2$. The slope of the tangent $T$ is $m_T = -\frac{1}{2}$. The line segment $F_1P'$ is perpendicular to the tangent $T$, so its slope is $m_{F_1P'} = 2$. The equation of the line $F_1P'$ passing through $F_1(-1, 0)$ with slope 2 is $y - 0 = 2(x - (-1)) \implies y = 2x + 2$. The equation of the tangent $T$ passing through $P(4, 6)$ with slope $-1/2$ is $y - 6 = -\frac{1}{2}(x - 4) \implies x + 2y - 16 = 0$. The intersection $P'(x', y')$ of $y=2x+2$ and $x+2y-16=0$ gives the foot of the perpendicular: $x + 2(2x+2) - 16 = 0 \implies 5x - 12 = 0 \implies x' = \frac{12}{5}$. $y' = 2(\frac{12}{5}) + 2 = \frac{24}{5} + \frac{10}{5} = \frac{34}{5}$. So, $P' = (\frac{12}{5}, \frac{34}{5})$.

6. (iii) Equation of the Auxiliary Circle: Center $C(1, 2)$ and radius $a=5$. $(x-1)^2 + (y-2)^2 = 5^2 \implies (x-1)^2 + (y-2)^2 = 25$.

7. (iv) Equations of the Directrices: The foot of the perpendicular from a focus to a tangent lies on the directrix for that focus. The slope of the major axis $F_1F_2$ is $m_{major} = \frac{4-0}{3-(-1)} = 1$. The directrices are perpendicular to the major axis, so their slope is $-1$.

   • Directrix for $F_2(3,4)$: Passes through $P(4, 6)$ with slope $-1$. $y - 6 = -1(x - 4) \implies x + y - 10 = 0$.
   • Directrix for $F_1(-1,0)$: Passes through $P'(\frac{12}{5}, \frac{34}{5})$ with slope $-1$. $y - \frac{34}{5} = -1(x - \frac{12}{5}) \implies 5y - 34 = -5x + 12 \implies 5x + 5y - 46 = 0$.